Solve

Solve
, 
Given , ![]() |
||
let ![]() |
= | p and q = ![]() |
| 2p + 6q | = | 2 ..... (1) |
| 3p – 2q | = | 4 ...... (2) |
Multiply eq(1) by 3 and eq(2) by 2, we get ![]() |
||
| q | = | – 11 |
| Substituting these value in equation (2) | ||
| 3p – 2(– 11) | = | 4 |
| 3p + 22 | = | 4 |
| 3p | = | – 18 |
| p | = | – 6 |
| Substituting p and q values, | ||
![]() |
= | (1/–6), = (1/–11) |
| (1/x) | = | (1/36), (1/y) = (1/121) |
| ∴ x | = | 36, y = 121 |
Solve (a + b)x + (a – b)y = a2 + b2, (a – b)(x + y) = a2 – b2
| Given (a + b)x + (a – b)y | = | a2 + b2 ..... (1) |
| (a – b)(x + y) | = | a2 – b2 ...... (2) |
| (a + b)x + (a – b)y – (a – b)x – (a – b)y | = | a2 + b2 – a2 + b2 |
| (a + b – a + b)x | = | 2b2 |
| ∴ x | = | b |
| (a + b)b + (a – b)y | = | a2 + b2 |
| ab + b2 + (a – b)y | = | a2 + b2 |
| (a – b)y | = | a2 – ab |
| (a – b)y | = | a(a – ab) |
| ∴ y | = | a |
3x + 7y = 25 , 5x – 5y = 25. find the xy + yx

| 5x | = | 25 + 5(1) = 25 + 5 |
| 5x | = | 30 |
| x | = | 30/5 = 6 |
| xy + yx | = | 61 + 16 |
| = | 6 + 1 = 7 |
2a + 3b = 17 , 2a + 2 – 3b + 1 = 5
let 2a be x and 3b be y 2a + 3b = 17 , (2a × 22) – (3b × 31) = 5
| x + y | = | 17 |
| (x × 4) – (y × 3) | = | 5 |
| 4x - 3y | = | 5 |
![]() |
||
| x | = | 17 – 9 = 8 |
| 2a | = | 8 |
| 2a | = | 23 |
| a | = | 3 |
| 3b | = | 9 |
| 3b | = | 32 |
| b | = | 2 |


